Assume that we are choosing n objects from $\\{a,\cdots,a,b,\cdots,b, c_{1},\cdots,c_{n}\\}$
where there are n a's and n b's.
If we choose k elements from $\\{a,\cdots,a,b,\cdots,b\\}$, there are $k+1$ ways to do this;
and then we still have to choose $n-k$ elements from $\\{c_1,\cdots,c_n\\}$,
which can be done in $\binom{n}{n-k}=\binom{n}{k}$ ways.
Therefore the total number of choices is given by $\displaystyle\sum_{k=0}^{n} (k+1)\binom{n}{k}$.
Since $\displaystyle(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^k$, $\;\;\;\;\;\displaystyle x(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^{k+1}$; so differentiating gives $(x+1)^{n-1}[(n+1)x+1]=\displaystyle\sum_{k=0}^{n} (k+1)\binom{n}{k}x^k$.
Now substituting $x=1$ gives that $\displaystyle\sum_{k=0}^{n} (k+1)\binom{n}{k}=2^{n-1}(n+2)$.
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Another way to evalutate this would be to split up the sum as $\displaystyle\sum_{k=0}^{n} k\binom{n}{k}+\sum_{k=0}^n \binom{n}{k}$ to get $n2^{n-1}+2^n=2^{n-1}(n+2)$.